Course notes (chap. 1 Number Theory, chap. 2 ... - McGill University

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1.4.5 ∗∗∗ Extracting Square Roots modulo p e If we have a solution r to r 2 ≡ a (mod p), how do we find a solution s to s 2 ≡ a (mod p e )fore>1 The chinese remainder theorem does not apply here. We have to figure things out in a different way.
First, consider the case e =2. Sincer 2 ≡ a (mod p), there exists an integer m =(r 2 − a)/p such that r 2 − a = mp. Supposethesolutionmodp 2 is of the form s = r + kp for some integer k. Let’sexpands 2 : and therefore s 2 =(r + kp) 2 = r 2 +2rkp +(kp) 2 = mp + a +2rkp +(kp) 2 s 2 ≡ a +(m +2rk) ∗ p (mod p 2 ). We find a solution s by making m +2rk amultipleofp so that (m +2rk) ∗ p ≡ 0 (mod p 2 ).
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COMP-547A /B Cryptography and Data
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Euclid Leonhard Euler Greatest Comm
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1.2.1 Fast modular exponentiation T
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1.2.2 GCD calculations and multipli
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EXAMPLE gcd(7,19) = gcd(19,7) = gcd
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EXAMPLE gcd(7,19) = gcd(1,0) = 1 x
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EXAMPLE 7x ≡ 5 (mod 9) 7 -1 ∙7x
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Finally, we know that a solution x
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1.3.1 Chinese Remainder Theorem Qin
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1.4 Quadratic Residues Quadratic re
Page 23 and 24: Properties ( 1 =+1 n) ( ) ab ( a )
Page 25 and 26: EXAMPLE Jacobi(527, 723) = -Jacobi(
Page 27 and 28: Pierre de Fermat 1.4.2 Fermat-Euler
Page 29 and 30: For prime numbers p ≡ 1(mod4)anda
Page 31 and 32: EXAMPLE p = 37, a = 16 z = 5 (5 +
Page 37 and 38: EXAMPLE n = 37*43 = 1591, a = 16
Page 39 and 40: Definition 1.7 (SQROOT) The square
Page 41 and 42: 1.5 Prime numbers If we want a rand
Page 43 and 44: EXAMPLE n = 37*43 = 1591, n-1 = 159
Page 45 and 46: EXAMPLE n = 1597, n-1 = 1596 = 399*
Page 47 and 48: Algorithm 1.6 ( Miller-Rabin prime(
Page 49 and 50: In August of 2002, Agrawal, Kayal,
Page 51 and 52: Remark: If n is a prime, then it is
Page 53 and 54: 3: Pick x ∈ R Z n ∗ 4: c i ←
Page 55 and 56: Examples F 2 =({0, 1}, ⊕, ∧). F
Page 57 and 58: Algorithm 2.1 ( Primitive(q) ) 1: L
Page 59 and 60: Factoring q − 1... The only effic
Page 61 and 62: EXAMPLE n = 10000000 S = [2710128,
Page 63 and 64: EXAMPLE n = 10000000 S = [4721693,
Page 65 and 66: We also have operations g(x) modh(x
Page 67 and 68: F 2 x +1 x 9 + x 4 +1 x 2 + x +1 x
Page 69 and 70: 2.3 General Fields Let p be a prime
Page 71 and 72: 10011×01110 = 01001 since (x 4 +x+
Page 73: EXTRA SLIDES
Page 77 and 78: Second, notice that the same exact
Page 79 and 80: Suppose Alice wants to distribute a
Page 81: Algorithm 2.6 ( Solve(x 1 ,x 2 ,...
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Course notes (chap. 1 Number Theory, chap. 2 ... - McGill University

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